# Normal subgroup

## Definition

Let [ilmath](G,\times)[/ilmath] be a group and [ilmath]H[/ilmath] a subgroup of [ilmath]G[/ilmath], we say [ilmath]H[/ilmath] is a normal subgroup[1] of [ilmath]G[/ilmath] if:

• $\forall x\in G[xH=Hx]$ where the [ilmath]xH[/ilmath] and [ilmath]Hx[/ilmath] are left and right cosets
• This is the sameas saying: [ilmath]\forall x\in G[xHx^{-1}=H][/ilmath]

According to Serge Lang[1] this is equivalent (that is say if and only if or [ilmath]\iff[/ilmath])

• [ilmath]H[/ilmath] is the kerel of some homomorphism of [ilmath]G[/ilmath] into some other group
This can be summed up as the following two statements:
1. The kernel of a homomorphism is a normal subgroup
2. Every normal subgroup is the kernel of some homomorphism

## Proof of claims

Claim 1: $\forall x\in G[xH=Hx]\iff\forall x\in G[xHx^{-1}=H]$

Proof of: $\forall x\in G[xH=Hx]\implies\forall x\in G[xHx^{-1}=H]$

Suppose that for whatever [ilmath]g\in G[/ilmath] we have that [ilmath]gH=Hg[/ilmath] - we wish to show that for any [ilmath]x\in G[xHx^{-1}=H][/ilmath]
Let [ilmath]x\in G[/ilmath] be given.
Recall that $X=Y\iff[X\subseteq Y\wedge X\supseteq Y]$ so we need to show:
1. $xHx^{-1}\subseteq H$
2. $xHx^{-1}\supseteq H$
Let us show 1:
Suppose $y\in xHx^{-1}$ we wish to show $\implies y\in H$ (that is $xHx^{-1}\subseteq H$)
$y\in xHx^{-1}\implies \exists h_1\in H:y=xh_1x^{-1}$
$\implies yx=xh_1$ - note that [ilmath]xh_1\in xH[/ilmath]
By hypothesis, $\forall g\in G[gH=Hg]$
So, as [ilmath]yx=xh_1\in xH[/ilmath] we see [ilmath]yx\in Hx[/ilmath]
This means $\exists h_2\in H$ such that $yx=h_2x$
Using the cancellation laws for groups we see that
$y=h_2$ as [ilmath]h_2\in H[/ilmath] we must have [ilmath]y\in H[/ilmath]
We have now shown that $[y\in xHx^{-1}\implies y\in H]\iff[xHx^{-1}\subseteq H]$
Now to show 2:
Suppose that [ilmath]y\in H[/ilmath] we wish to show that $\implies y\in xHx^{-1}$ (that is $H\subseteq xHx^{-1}$)
Note that [ilmath]yx\in Hx[/ilmath] by definition of [ilmath]Hx[/ilmath]
By hypothesis [ilmath]Hx=xH[/ilmath] so
we see that [ilmath]yx\in xH[/ilmath]
this means [ilmath]\exists h_1\in H[yx=xh_1][/ilmath]
and this means [ilmath]y=xh_1x^{-1}[/ilmath]
such a [ilmath]h_1[/ilmath] existing is the very definition of [ilmath]xh_1x^{-1}\in xHx^{-1} [/ilmath]
thus [ilmath]y\in xHx^{-1} [/ilmath]
We have now shown that $[y\in H\implies y\in xHx^{-1}]\implies[H\subseteq xHx^{-1}]$
Combining this we hve shown that [ilmath]\forall x\in G[xH=Hx]\implies\forall x\in G[xHx^{-1}=H][/ilmath]

Next:

Proof of: $\forall x\in G[xHx^{-1}=H]\implies\forall x\in G[xH=Hx]$

TODO: Simple proof

Claim 2: The kernel of a homomorphism is a normal subgroup

We wish to show that given a homomorphism [ilmath]f:G\rightarrow X[/ilmath] (where [ilmath]X[/ilmath] is some group) that the kernel of [ilmath]f[/ilmath], [ilmath]H[/ilmath] is normal. Which is to say that:

• $\forall x\in G[xHx^{-1}=H]$ (which is $\forall x\in G[x\text{Ker}(f)x^{-1}=\text{Ker}(f)]$ )

Proof that $\forall x\in G[xHx^{-1}\subseteq H]$

Let [ilmath]x\in G[/ilmath] be given
Let [ilmath]y\in xHx^{-1} [/ilmath] be given
Then $\exists h_1\in H:y=xh_1x^{-1}$
$f(y)=f(xh_1x^{-1})=f(x)f(h_1)f(x^{-1})$
But [ilmath]H[/ilmath] is the kernel of [ilmath]f[/ilmath] so [ilmath]f(h_1)=e[/ilmath] where [ilmath]e[/ilmath] is the identity of [ilmath]X[/ilmath]
$f(y)=f(x)ef(x^{-1})$
It is a property of homomorphisms that [ilmath]f(x^{-1})=(f(x))^{-1}[/ilmath]
$f(y)=f(x)f(x^{-1})=f(x)f(x)^{-1}=e$
Thus [ilmath]y\in\text{Ker}(f)=H[/ilmath]
So we see that [ilmath]xHx^{-1}\subseteq H[/ilmath]

Proof that $\forall x\in G[H\subseteq xHx^{-1}]$

As before, let [ilmath]x\in G[/ilmath] be given.
From above we know that $\forall x\in G[xHx^{-1}\subseteq H]$, using this we see that $x^{-1}Hx\subseteq H$
Let [ilmath]y\in H[/ilmath] be given (we will show that then [ilmath]y\in xHx^{-1} [/ilmath])
We know that [ilmath]x^{-1}Hx\subseteq H[/ilmath], as [ilmath]y\in H[/ilmath] we know that [ilmath]x^{-1}yx\in H[/ilmath]
This means $\exists h\in H[x^{-1}yx=h]$
$\implies yx=xh$
$\implies y=xhx^{-1}$
Thus $y\in xHx^{-1}$
We have shown that $[y\in H\implies y\in xHx^{-1}]\iff[H\subseteq xHx^{-1}]$

We have shown that $\forall x\in G[xHx^{-1}\subseteq H\wedge H\subseteq xHx^{-1}]$, which is exactly:

• Given a group homomorphism [ilmath]f:G\rightarrow X[/ilmath] where [ilmath]\text{Ker}(f)=H[/ilmath], we have shown that [ilmath]\forall x\in G[H=xHx^{-1}][/ilmath] which is the first definition of a normal subgroup

Claim 3: Every normal subgroup is the kernel of some homomorphism

## References

1. Undergraduate Algebra - Serge Lang